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Gravity would only be weaker at the equator if mass is not the only thing that produces gravity or if there is dense enough matter near, or at, the center to offset the additional volume of mass that creates more space between the equator and center.

Centrifugal force is an outside force. Just because I blow air upwards on an object, changing it's fall rate, doesn't mean gravitational force has changed.

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    $\begingroup$ Are you getting confused with apparent weight? $\endgroup$ Commented Jun 26, 2022 at 21:05

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There are two things that could give meaning to your question.

First, there's the phenomena mentioned in rhomaios's answer: Earth's radius is slightly higher at the equator, so objects at the surface of the Earth there feel a slightly weaker gravitational force.

Second, there's apparent gravity. Let's go into more details, because it seems to be what you're after (according to the title).

In a reasonably small area at the surface of the Earth, gravitational force is well approximated as a force such that:

  • it has constant direction
  • it's proportional to mass, with a proportionality coefficient that is constant.

$$\vec{P} =m\vec{g} \simeq -mg\vec{e}_z \quad\text{with}\quad g=G\frac{M_T}{R_T^2}$$

There's also the centrifugal force and, also in a small area at the surface of the Earth, it is correctly approximated in a very similar way:

  • constant direction
  • proportional to mass, with a proportionality coefficient that is constant at a given latitude.

$$\vec{F}_c=m\omega^2R_T\cos(\lambda)\vec{u}$$

with $\lambda$ the latitude and $\vec{u}$ a unitary vector perpendicular to the Earth's rotation axis.

At the equator, $\lambda=0$ and $\vec{u}_z=\vec{u}$ so both forces are collinear with opposite directions. Their sum is usually called the apparent weight: $$\vec{F}=m(-g+\omega^2R_T)\vec{u}_z$$

Numerically the second (centrifugal) term is much smaller than the first (gravitational). So someone stepping on a scale will be slightly "lighter", because the scale cannot separate both forces.

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I have never seen anyone claim that gravity is weaker due to the centrifugal force. Gravity is (extremely slightly) weaker at the equator because the earth is not a perfect sphere, hence the distance from the equator to its center is somewhat larger than the distance to other points on the surface (with the poles being the two closest points). Since Newtonian gravity diminishes with distance as per the inverse square law, the gravitational interaction between the earth and some other object on it is weakest at the equator.

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  • $\begingroup$ Gravity is not weaker but your weight (force of you on the earth) is weaker if you are fixed in a rotating (non-inertial) frame due to the centrifugal force. See for example. Fowles, Analytical Mechanics, or Symon, Mechanics. $\endgroup$
    – John Darby
    Commented Jun 26, 2022 at 21:57
  • $\begingroup$ Indeed, but the OP was referring to gravity. They might in fact be referring to apparent weight, as both you and Miyase suggest, or (more likely IMO) they somehow conflated the two notions (weaker gravity at the equator and lower apparent weight on the equator also). $\endgroup$
    – rhomaios
    Commented Jun 26, 2022 at 22:01
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    $\begingroup$ Yes. What we all have provided should clarity this. $\endgroup$
    – John Darby
    Commented Jun 26, 2022 at 22:07
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The answer by @Miyase is correct. A simple discussion of the effect of being fixed in a non-inertial (rotating) frame follows. Hope this helps.

Your "weight" is the force you exert on the earth when you are at a fixed position on the earth.

Here, to address the centrifugal force, I assume the earth is a sphere so the force of gravity on you at it's surface is constant regardless of your position (pole, equator, etc.)

For you at fixed at the equator, relative to a fixed inertial system you have a centripetal acceleration due to your circular motion; that centripetal force is the force of gravity (inward) on you minus the force of the earth (outward) on you. At the pole you have no circular motion hence the force of gravity is equal in magnitude to the force of the earth on you.

For you at fixed at the equator, you are fixed in a non-inertial (rotating) frame, and in this frame an outward centrifugal force (a fictitious force) acts on you. For you at rest in this frame, the centrifugal force outward plus the force of the earth outward equals the force of gravity inward to keep you at rest. So, the magnitude of the force of the earth on you = magnitude of force of gravity minus magnitude of centrifugal force; the centrifugal force reduces the force of the earth on you. The force of the earth on you is equal in magnitude to the force of you on the earth, which is your weight, so you weigh less due to the centrifugal force.

If you are moving relative to the earth, another fictitious force, the Coriolis force, is present. This force is responsible for the "drift" of a projectile.

Physics mechanics textbooks address fictitious forces in non-inertial (accelerating) frames of reference in some detail, and you can also find useful information online.

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  • $\begingroup$ Since you mention the Coriolis force, one additional thing I didn't include in my answer: this force is never added to apparent weight for (at least) two reasons: 1) it depends on velocity so by studying your motion you can separate it from the gravitational and centrifugal forces; 2) its direction is often close to horizontal, which is also very different from the other two. $\endgroup$
    – Miyase
    Commented Jun 26, 2022 at 22:05
  • $\begingroup$ Yes. Good comment. $\endgroup$
    – John Darby
    Commented Jun 26, 2022 at 22:06

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